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According to the figure, CA is tangent to the circle, centre O, at A. ABT and POT are straight lines.

Question:

Given that BT is equal to the radius of the circle, prove that:

  1. $\angle ABP = 3 \angle OBP$
  2. $\angle POA = 3 \angle BOT$
  3. $\angle OBT = 2 (\angle BTO + \angle CAB)$

enter image description here

My attempt:

  • $\angle AOP = 2 \angle ABP. $
  • $\angle OBP = \angle OPB.$
  • $\angle CAB = \angle APO.$
  • $\angle BOA = 2 \angle APB.$
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  • $\begingroup$ What is the question? $\endgroup$ May 26 '20 at 6:58
  • $\begingroup$ prove that: 1. Angle ABP = 3 time of Angle OBP 2. Angle POA = 3 times of Angle BOT 3. Angle OBT = 2 times of (Angle BTO + Angle CAB) $\endgroup$
    – IM_LOST
    May 26 '20 at 6:59
  • $\begingroup$ Have you made any attempt? Please add an attempt so we can help you with where you might've got stuck $\endgroup$ May 26 '20 at 6:59
  • $\begingroup$ Yes, I have edited the original post to include my attempts. $\endgroup$
    – IM_LOST
    May 26 '20 at 7:10
  • $\begingroup$ Use the isosceles properties of BOT, AOB and AOP to relate the angles along with the AOP = 2* ABP. That should solve all of them $\endgroup$ May 26 '20 at 7:23
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a):

$BOT=2PBO$

$ABO=2BOT=4PBO$

$ABP=ABO-PBO=4PBO-PBI=3PBO$

b):

$ABP+\frac{1}{3}ABP=2BOT$

$\frac{4}{3}ABP=2BOT$

$ABP=\frac{1}{2}POA$

$\frac{4}{3}(\frac{1}{2}POA)=2BOT$

Therefore:

$POA=3BOT$

c):

$ OBT=AOB+OBA$

$AOB=2ABP=2CAB$

$OBA=2BTO$

Therefore:

$OBT 2(CAB+BTO)$

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  1. $$\measuredangle ABP=\measuredangle OPB+\measuredangle OTB=\measuredangle OPB+\measuredangle TOB=$$ $$=\measuredangle OPB+\measuredangle OPB+\measuredangle OBP=3\measuredangle OBP.$$
  2. $$\measuredangle POA=\measuredangle OTB+\measuredangle OAB=\measuredangle OTB+\measuredangle OBA=$$ $$=\measuredangle OTB+\measuredangle OTB+\measuredangle BOT=3\measuredangle BOT.$$
  3. $$\measuredangle OBT=\measuredangle OAB+\measuredangle AOB=\measuredangle ABO+2\measuredangle CAB=$$ $$=\measuredangle BTO+\measuredangle BOT+2\measuredangle CAB=2(\measuredangle BTO+\measuredangle CAB).$$
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